How many grams of ammonia can be produced from reacting a 450 L sample of nitrogen gas at a temperature of 450 K and a pressure of 300 atm?

The mass of ammonia is 124457.96 g which can be produced from reacting a 450 L sample of nitrogen gas at a temperature of 450 K and a pressure of 300 atm.

Explanation:

Given: Volume = 450 L

Temperature = 450 K

Pressure = 300 atm

The reaction equation is as follows.

Here, 1 mole of nitrogen reacts to give 2 moles of ammonia.

From the given data, moles of nitrogen are calculated as follows.

PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into the above formula as follows.

For 3654.08 moles of nitrogen, the moles of ammonia produced is as follows.

Therefore, mass of ammonia (molar mass = 17.03 g/mol) is calculated as follows.

Thus, we can conclude that the mass of ammonia is 124457.96 g which can be produced from reacting a 450 L sample of nitrogen gas at a temperature of 450 K and a pressure of 300 atm.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

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